Add definite assignment checks for interacting with `noUncheckedIndexAccess`

[chriskrycho]

Suggestion

🔍 Search Terms

noUncheckedIndexAccess, narrowing, assignment, definite assignment, control flow

✅ Viability Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

⭐ Suggestion

It'd be great if noUncheckedIndexAccess understood when you have just initialized a given index in terms of control flow, the same way that TS can see that a standalone optional binding is definitely initialized.

📃 Motivating Example

const stringNumDict: Record<string, number> = Object.create(null);

function isNonZero(key: string, dict: Record<string, number>): boolean {
  if (!dict[key]) {
    dict[key] = 0;
  }

  return dict[key] > 0;
}

Here, isNonZero has to insert a non-null assertion operator on the final line of the function to type check. (Mind, this is a bad pattern; it should be initialized elsewhere. I have seen this come up in the real world in things like a reduce building a nested map, though.) For a standalone value with the same semantics as dict[key], i.e. number | undefined, the assignment informs TS that it is no longer | undefined, so the final check works.

💻 Use Cases

In the simplest cases (with value types), there is no workaround other than using a non-null assertion operator or an explicit cast. In cases with reference types (e.g. assigning an object if the value isn't set), doing a check like this can suffice as a workaround:

let nested = dict[key];
if (!nested) {
  nested = dict[key] = {};
}

nested.otherStuff = ...;

[mkantor]

This has been fixed, possibly by #57847. In versions of TypeScript since 5.5, the motivating example no longer has an error.

[chriskrycho]

So it has—thanks for flagging that! Closing this as completed. 🎉